a)
(x - 2)(x + 2)(x - p)
= (x² - 4)(x - p)
= x³ - px² - 4x + 4p

Multiply by 5 ,
f(x) = 5x³ - 5px² - 20x + 20p

b)

f(-1) = -30

5(-1)³ - 5p(-1)² - 20(-1) + 20p = -30
-5 - 5p + 20 + 20p = -30
15p = -45
p = -3

c)

2x + 1 = 2(x + ½)

Remainder = f(-½)

= 5 (-½)³ - 5p(-½)² - 20(-½) + 20p
= -⅝ - 5/4 p + 10 + 20p
= 9⅜ + 18¾ p

When p = -3,

Remainder = 9⅜ + 18¾(-3)
= -46⅞


d)

From a) we know that f(x)
= 5x³ + 15x² - 20x - 60
= 5(x³ + 3x² - 4x - 12) ①
= 5(x - 2)(x + 2)(x - p)
= 5(x - 2)(x + 2)(x + 3) = 0

e^(4x) + 3e^(2x) - 4 = 12e^(-2x)
Multiply by e^(2x),

e^(6x) + 3e^(4x) - 4e^(2x) = 12

e^(6x) + 3e^(4x) - 4e^(2x) - 12 = 0

(e^(3x))² + 3(e^(2x))² - 4e^(2x) - 12 = 0

We can see that this is analogous to ① where we replace x by e^(2x).

So we rewrite it as :

(e^(2x) - 2)(e^(2x) + 2)(e^(2x) + 3) = 0

e^(2x) = 2 or e^(2x) = -2 or e^(2x) = -3

(The latter two are rejected as e^(2x) > 0 for all real x. In other words the exponential is always positive for real values of x)

So e^(2x) = 2

2x = ln2
x = ½ln2

= ln(2)^½
= ln√2

Thanks . clearly understood

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